A projectile fired with a speed of v=60m/s at an angle of 60°. a second projectile is then fired with the same initial speed 0.5s later. Find the angle #theta# of the second projectile so that they will collide and what position (x, y) will it occur?

1 Answer
Narad T.
Jun 12, 2018

The angle is #theta=57.58^@# and the position is #(222.6, 116.1)#

Explanation:

The horizontal distance travelled is

#{(x=vcos60t),(x=vcostheta(t-0.5)):}#

Therefore,

#60*1/2*t=60costheta(t-0.5)#

#30t=60(t-0.5)costheta#

#t=2tcostheta-costheta#

#t(2costheta-1)=costheta#

#t=costheta/(2costheta-1)#

The vertical distance travelled is

#{(y=vsin60*t-1/2g t^2=52t-4.9t^2),(y=vsintheta(t-0.5)-1/2g(t-0.5)^2=60tsintheta-30sintheta-4.9(t-0.5)^2):}#

#{(y=52t-4.9t^2),(y=60tsintheta-30sintheta-4.9t^2+4.9t-1.23):}#

Therefore,

#52t-4.9t^2=60tsintheta-30sintheta-4.9t^2+4.9t-1.23#

#60tsintheta-30sintheta+4.9t-52t-1.23=0#

#t(60sintheta-47.1)=30sintheta+1.23#

#t=(30sintheta+1.23)/(60sintheta-47.1)#

The #t#'s must be equal

#(30sintheta+1.23)/(60sintheta-47.1)=costheta/(2costheta-1)#

#(30sintheta+1.23)(2costheta-1)=costheta(60sintheta-47.1)#

#60sinthetacostheta-30sintheta+2.46costheta-1.23=60sintheta costheta-47.1costheta#

#-30sintheta+49.56costheta=1.23#

#49.56costheta-30sintheta=1.23#

Solving for #theta#

#rsin(alpha-theta)=rsinalphacostheta-rsinthetacosalpha#

#{(rcosalpha=-30),(rsinalpha=49.56):}#

#<=>#, #r^2=sqrt(30^2+49.56^2)=57.93#

#{(cosalpha=-30/57.93=-0.558),(sinalpha=49.56/57.93=0.83):}#

#alpha=58.8^@#

Therefore,

#57.93sin(58.8-theta)=1.23#

#sin(58.8-theta)=1.23/57.93=0.021#

#58.8-theta=1.22#

#theta=58.8-1.22=57.58^@#

And

#t=cos(57.6)/(2cos(57.6)-1)=7.42s#

#x=60cos(57.58)(7.42-0.5)=222.6m#

#y=52*7.42-4.9*7.42^2=116.1m#

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